1. Chapter 22 Gauss’s Law
The Study guide is posted online under the homework section ,
Your exam is on March 6
Chapter 22 opener. Gauss’s law is an elegant relation between electric charge and electric field. It is more general than Coulomb’s law. Gauss’s law involves an integral of the electric field E at each point on a closed surface. The surface is only imaginary, and we choose the shape and placement of the surface so that we can figure out the integral. In this drawing, two different surfaces are shown, both enclosing a point charge Q. Gauss’s law states that the product E·dA, where dA is an infinitesimal area of the surface, integrated over the entire surface, equals the charge enclosed by the surface Qencl divided by ε0. Both surfaces here enclose the same charge Q. Hence ∫E·dA will give the same result for both surfaces.

2. 22-2 Gauss’s Law
The net number of field lines through the surface is proportional to the charge enclosed, and also to the flux, giving Gauss’s law:
This can be used to find the electric field in situations with a high degree of symmetry.

3. 22-2 Gauss’s Law-Problem 7
Problem 7.(II): In Fig. 22–27, two objects O1 and O2, have charges μC and μC respectively, and a third object, O3 is electrically neutral. (a) What is the electric flux through the surface A1 that encloses all the three objects? (b) What is the electric flux through the surface A2 that encloses the third object only?

4. 22-2 Gauss’s Law For a point charge, Therefore,
Figure A single point charge Q at the center of an imaginary sphere of radius r (our “gaussian surface”—that is, the closed surface we choose to use for applying Gauss’s law in this case).
Solving for E gives the result we expect from Coulomb’s law:

5. 22-2 Gauss’s Law
Using Coulomb’s law to evaluate the integral of the field of a point charge over the surface of a sphere surrounding the charge gives:
Figure A single point charge surrounded by a spherical surface, A1, and an irregular surface,A2.
Looking at the arbitrarily shaped surface A2, we see that the same flux passes through it as passes through A1. Therefore, this result should be valid for any closed surface.

6. 22-2 Gauss’s Law
Finally, if a Gaussian surface encloses several point charges, the superposition principle shows that:
Therefore, Gauss’s law is valid for any charge distribution. Note, however, that it only refers to the field due to charges within the gaussian surface – charges outside the surface will also create fields.

7. 22-2 Gauss’s Law Conceptual Example 22-2: Flux from Gauss’s law.
Consider the two Gaussian surfaces, A1 and A2, as shown. The only charge present is the charge Q at the center of surface A1. What is the net flux through each surface, A1 and A2?
Solution: The net flux through A1 is Q/ε0, as it encloses charge Q. The net flux through surface A2 is zero, even though the field is not zero on the surface.

8. 22-3 Applications of Gauss’s Law
Example 22-3: Spherical conductor.
A thin spherical shell of radius r0 possesses a total net charge Q that is uniformly distributed on it. Determine the electric field at points (a) outside the shell, and (b) within the shell. (c) What if the conductor were a solid sphere?
Figure Cross-sectional drawing of a thin spherical shell of radius r0 carrying a net charge Q uniformly distributed. A1 and A2 represent two gaussian surfaces we use to determine Example 22–3.
Solution: a. The gaussian surface A1, outside the shell, encloses the charge Q. We know the field must be radial, so E = Q/(4πε0r2).
b. The gaussian surface A2, inside the shell, encloses no charge; therefore the field must be zero.
c. All the excess charge on a conductor resides on its surface, so these answers hold for a solid sphere as well.

9. 22-3 Applications of Gauss’s Law
Example 22-4: Solid sphere of charge.
An electric charge Q is distributed uniformly throughout a nonconducting sphere of radius r0. Determine the electric field (a) outside the sphere (r > r0) and (b) inside the sphere (r < r0).
Solution: a. Outside the sphere, a gaussian surface encloses the total charge Q. Therefore, E = Q/(4πε0r2).
b. Within the sphere, a spherical gaussian surface encloses a fraction of the charge Qr3/r03 (the ratio of the volumes, as the charge density is constant). Integrating and solving for the field gives E = Qr/(4πε0r03).

10. 22-3 Applications of Gauss’s Law
Example 22-5: Nonuniformly charged solid sphere.
Suppose the charge density of a solid sphere is given by ρE = αr2, where α is a constant. (a) Find α in terms of the total charge Q on the sphere and its radius r0. (b) Find the electric field as a function of r inside the sphere.
Solution: a. Consider the sphere to be made of a series of spherical shells, each of radius r and thickness dr. The volume of each is dV = 4πr2 dr. To find the total charge: Q = ∫ρE dV = 4παr05/5, giving α = 5Q/4πr05.
b. The charge enclosed in a sphere of radius r will be Qr5/r05. Gauss’s law then gives E = Qr3/4πε0r05.